[Proof of Theorem REF ]
Let \(u_1\ge u_2\ge \ldots \ge u_n\)  denote the eigenvalues of \(K(X,X))\) . Since \(K(X,X)\)  is non-negative definite, we have \(u_i\ge 0\)  for \(1\le i\le n\) . Note that \(\sup _{\mathbf {x}\in \mathcal {X}}K(\mathbf {x},\mathbf {x})<\infty \)  as \(K\)  is a continuous bivariate function on a compact support \(\mathcal {X}\times \mathcal {X}\) . Then we have \(\sum _{i=1}^{n}u_i=\mathrm {tr}(K(X,X))\lesssim n\) .
Let \(\mathbf {f}=(f(X_1),\ldots ,f(X_n))^T\) . The MMLE \(\hat{\sigma }_n^2\)  is a quadratic form in \(Y\) . Hence, in view of the well known formula for the expectation of quadratic forms (cf. Theorem 11.19 in [1]}), we obtain
\(\mathbb {E}(\hat{\sigma }^2_n)=\lambda \sigma _0^2\mathrm {tr}([K(X,X)+n\lambda \mathbf {I}_n]^{-1})+\lambda \mathbf {f}^T[K(X,X)+n\lambda \mathbf {I}_n]^{-1}\mathbf {f}.\) 
